A place kicker kicks a football with a velocity of 30.0

Can and object have a negative position and a positive velocity? Or vice versa, a positive position and a negative velocity? Exp

zavuch27 [327]

Imagine a ball is moving on the following horizontal line.

. . . . . . . . . . . . . . . . . . . O. . . . . . . . . . . . . . . . . .

Take right as positive. O is the starting point of the ball. Denote the ball by o.

. . . . . . . . . . . . . . . . . . . O. . . . . . . ... . . o . . . . . .

Assume the ball is moving to the right. It has positive displacement since it is on the right of O, and positive velocity since its positive displacement is increasing.

.ñ

. . . . . . . . . . . . . . . . . . . O. . . . o . . . . . . . . . . . . .

Now the ball is returning to O. It still has positive displacement since its current position is still on the right of O. However, its velocity is negative since its positive displacement is decreasing and the direction of the velocity vector points left, which is the negative side.

By now you should be able to come up with a scenario where the ball has negative displacement and positive velocity.

You can observe the same phenomenon in daily life. Say, as a stretched spring bounces to its starting position, if we let the returning direction be positive, the string has negative displacement since it is on the negative direction, but has positive velocity. Bungee jump can also used to illustrate the phenomenon.

4 0

2 years ago

When you hit a .27 kg volleyball the contact time is 50 ms and the average force is 125 N. If you serve the volleyball (from res

yulyashka [42]

(A) P(v) = 0.135v

(B) P(h) = 0.234v

<u>Explanation:</u>

Given-

Mass of the ball, m = 0.27kg

Force, F = 125N

angle of projection, θ = 30°

Let v be the velocity of the ball.

A) vertical component of the momentum of the volleyball

We know,

P(vertical) = mvsinθ

P(V) = 0.27 X v X sin 30°

P(V) = 0.27 X v X 0.5

P(V) = 0.135v

B) horizontal component of the momentum of the volleyball

We know,

P(Horizontal) = mvcosθ

P(h) = 0.27 X v X cos 30°

P(h) = 0.27 X v X 0.866

P(h) = 0.234v

8 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$